WEG-curso-dt-5-caracteristicas-e-especificacoes-de-geradores-artigo-tecnico-portugues-br
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DT-5 - Características e Especificações <strong>de</strong> Geradores<<strong>br</strong> />
II.1 - Contribuição individual do motor <strong>de</strong><<strong>br</strong> />
75cv (IN=87,5A - IP=647,5A).<<strong>br</strong> />
Supondo uma queda <strong>de</strong> tensão inicial <strong>de</strong> 15% e<<strong>br</strong> />
utilizando chave compensadora no TAP 65%:<<strong>br</strong> />
(1-0,15) x 0,65 = 0,85 x 0,65 = 0,55<<strong>br</strong> />
Então, da Fig. 5.3.5: K1=0,45<<strong>br</strong> />
Corrente <strong>de</strong> partida do motor referida ao primário<<strong>br</strong> />
da chave compensadora (terminais do gerador):<<strong>br</strong> />
IPmotor<<strong>br</strong> />
65%ref.= 647,5 .0,45 .0,65<<strong>br</strong> />
IPmotor<<strong>br</strong> />
65%ref.= 189A<<strong>br</strong> />
Relação IP/IN nos terminais do gerador:<<strong>br</strong> />
IP<<strong>br</strong> />
motor 65% ref.<<strong>br</strong> />
Ig<<strong>br</strong> />
189<<strong>br</strong> />
=<<strong>br</strong> />
383 = 0,493<<strong>br</strong> />
A queda <strong>de</strong> tensão que ocorrerá, consi<strong>de</strong>rando<<strong>br</strong> />
somente a partida do motor <strong>de</strong> 75cv será:<<strong>br</strong> />
0,1839.0,493<<strong>br</strong> />
V =<<strong>br</strong> />
.100<<strong>br</strong> />
1+ [ 0,1839.0,493 ]<<strong>br</strong> />
V = 8, 31%<<strong>br</strong> />
II.2 - Contribuição dos motores <strong>de</strong> 100 e<<strong>br</strong> />
25cv quando da partida do motor <strong>de</strong><<strong>br</strong> />
75cv:<<strong>br</strong> />
NOTA: O processo <strong>de</strong> cálculo é iterativo e segue o<<strong>br</strong> />
roteiro mostrado abaixo:<<strong>br</strong> />
II.2.1 - Valor suposto <strong>de</strong> queda = 15%.<<strong>br</strong> />
Do gráfico da fig. 5.3.4, obtemos o<<strong>br</strong> />
incremento <strong>de</strong> corrente dos motores<<strong>br</strong> />
em carga para uma redução <strong>de</strong> tensão<<strong>br</strong> />
em seus terminais <strong>de</strong> 15%. Para o<<strong>br</strong> />
caso em questão temos i = 26% (=<<strong>br</strong> />
0,26).<<strong>br</strong> />
Logo, os acréscimos <strong>de</strong> corrente dos motores serão:<<strong>br</strong> />
Motor <strong>de</strong> 100cv (IN=120A para 440V):<<strong>br</strong> />
Acréscimo = i . 120 = 0,26 . 120<<strong>br</strong> />
Acréscimo = 31,2A<<strong>br</strong> />
i do motor <strong>de</strong> 100cv:<<strong>br</strong> />
Acréscimo 31,2<<strong>br</strong> />
i (M100)= =<<strong>br</strong> />
Ig 383<<strong>br</strong> />
i(M100) = 0,0815<<strong>br</strong> />
Motor <strong>de</strong> 25cv (IN=31,5A para 440V):<<strong>br</strong> />
Acréscimo = i . 31,5 = 0,26 . 31,5<<strong>br</strong> />
Acréscimo = 8,2 A<<strong>br</strong> />
i do motor <strong>de</strong> 25cv:<<strong>br</strong> />
8,2<<strong>br</strong> />
i (M25)=<<strong>br</strong> />
383 = 0,0214<<strong>br</strong> />
Cálculo do IP/IN total:<<strong>br</strong> />
IP IP<<strong>br</strong> />
= (M75)+ i(M100)+ i(M25)<<strong>br</strong> />
IN Ig<<strong>br</strong> />
IP<<strong>br</strong> />
= 0 ,493+<<strong>br</strong> />
0,0815+<<strong>br</strong> />
0,0214<<strong>br</strong> />
IN<<strong>br</strong> />
IP<<strong>br</strong> />
= 0,<<strong>br</strong> />
IN<<strong>br</strong> />
IP/IN total: 5959<<strong>br</strong> />
0,1839.0,5959<<strong>br</strong> />
V =<<strong>br</strong> />
.100<<strong>br</strong> />
1+ [ 0,1839.0,5959 ]<<strong>br</strong> />
V = 9, 88%<<strong>br</strong> />
Como supomos V = 15% e resultou numa queda<<strong>br</strong> />
<strong>de</strong> 9,88% refaremos o cálculo:<<strong>br</strong> />
II.2.2 - Admitindo agora a queda <strong>de</strong> tensão V =<<strong>br</strong> />
9,88%, da fig. 7.3.4: i = 15% (=0,15)<<strong>br</strong> />
Acréscimos <strong>de</strong> corrente nos motores:<<strong>br</strong> />
(1-0,10) x 0,65 = 0,90 x 0,65 = 0,585 ; K1=0,49<<strong>br</strong> />
motor 49 .0,65 = 206,22A<<strong>br</strong> />
IP 65%ref.= 647,5 .0,<<strong>br</strong> />
IP 206,22<<strong>br</strong> />
(M75) 0,5384<<strong>br</strong> />
Ig 383<<strong>br</strong> />
IP IP<<strong>br</strong> />
=<<strong>br</strong> />
IN Ig<<strong>br</strong> />
(M75)+ i(M100)+ i(M25)<<strong>br</strong> />
IP 120.0,15<<strong>br</strong> />
31,5.0,15 <<strong>br</strong> />
= 0 ,5384+<<strong>br</strong> />
+ <<strong>br</strong> />
IN 383 383 <<strong>br</strong> />
IP<<strong>br</strong> />
= 0,5384<<strong>br</strong> />
0,047 0,01234<<strong>br</strong> />
0.5977<<strong>br</strong> />
IN<<strong>br</strong> />
0,1839 . 0,5977<<strong>br</strong> />
V =<<strong>br</strong> />
.100<<strong>br</strong> />
1+ [ 0,1839 . 0,5977]<<strong>br</strong> />
V = 9, 90%<<strong>br</strong> />
Logo, V estipulado V calculado.<<strong>br</strong> />
Po<strong>de</strong>remos encerrar o cálculo.<<strong>br</strong> />
Indicação: Recalcular queda <strong>de</strong> tensão com mo<strong>de</strong>lo<<strong>br</strong> />
inferior, para verificar se é possível escolha mais<<strong>br</strong> />
econômica.<<strong>br</strong> />
CONCLUSÃO: Po<strong>de</strong>mos observar que, para este<<strong>br</strong> />
caso, a contribuição dos motores já em<<strong>br</strong> />
funcionamento não causou um acréscimo muito<<strong>br</strong> />
significativo na queda <strong>de</strong> tensão geral.<<strong>br</strong> />
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